(a) The thermal efficiency of the cycle.
The thermal efficiency of the cycle is given by
NThermal= net work/ heat supplied
NThermal= WNet/QH
=2026.08/4547.82= 44.55%
(b) The backwork ratio:
compressor work/ turbine work
=1760.76/3786.84= 0.465
(c) The net power developed:
WNet= WT-WC
WNet= 3786.84- 1760.76= 2026.08 kW
Answer:
Air cooling.
Explanation:
Low power motors are supposed to be low cost, and they dissipate little heat. Therefore a low cost solution is ideal.
Air cooling can be achieved with very little cost. Fins can be added to a cast motor casing and a fan can be places on the shaft to use a small amount of the motor power to move air to cool it.
Given Information:
Resistance = R = 2 Ω
Capacitance = C = 4 F
Required Information:
Time constant = τ = ?
Answer:
τ = 8 seconds
Explanation:
one time constant τ is the amount of time taken when the capacitor chargers up to approximately 63 % of its maximum possible voltage.
τ = RC
τ = 2*4
τ = 8 seconds
So that means after 1 τ (8 seconds) the capacitor voltage will reach approximately 63.2 % of its maximum possible voltage.
I’m sorry but I can’t read all of it maybe you should write it in type so we can read it.
Answer:
Pull, aim, squeeze, sweep
